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Author: Subject: Reduction of KMnO4
Madchemyst
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[*] posted on 22-4-2015 at 05:46
Reduction of KMnO4


OK, I understand how to balance equations, I understand how to determine oxidation states. What I am very confused about and my head hurts trying to understand is how do you know all the products that are produced and that something was reduced. In doing some research and reading on chemguide, for example in a reaction of KMnO4 + H2O2 it states that "you are told or you know that Mn(VII) ions turns into Mn(II) ions". How do you know that? Is it simply a guess looking at Electronegitivity and the fact the H2O2 is a strong oxidizer, so Mn must be reduced? Also when you look at someone else's reaction sometime how some of the products are derived is not clear. For example if HNO3 is used as a reagent and one of the products is a NO3 it will also list NO as a product with O2 or someting else. I understand if I am looking at a complete reaction that is balanced but how do you figure all the products that will be produced simply from the reagents? Hope what I am asking makes sense and any help would be much appriciated so my head can stop hurting
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byko3y
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[*] posted on 22-4-2015 at 06:07


Actually it's Mn(VI) and Mn(III), two most table states.
Chemistry is all about practice. You can't know nothing until you test it.
Reduction-oxidation potential can be measured, but they are different at different ph.
The only way nitric acid can be dehydrated is 2HNO3 -> H2O + N2O5 , but the N2O5 contains too much oxygen (which otherwise is stabilized by hydrogen), so it decomposes to NO2+O2, as well as easily hydrolises back N2O5 + H2O -> 2HNO3. For this reason you need a dehydration agent to make N2O5 ( 6HNO3 + P2O5 -> 3N2O5 + 2H3PO4 ). Once again: anything else is not a dehydration, but a decomposition of nitric acid, not a dehydration.
Why different compounds have different stability? Well, maybe because the change in one electron of the electron configuration changes the properties of the whole molecule.

[Edited on 22-4-2015 by byko3y]
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[*] posted on 22-4-2015 at 06:10


Someone else will give you the technical information better than I can and so I won't try.
What I will say is that some of this is a matter of experience. Oxidation states (inasmuch as they are real entities) are based on observations of real reactions and there is a body of experience that tells us what is likely to happen in any given scenario.
Of course there are ways of predicting as well and a table of standard reduction potentials is a tool here. However, kinetic considerations come into it as well and these things are not always simple. I will refrain from giving an explanation since it is late where I am and I have a habit of screwing up redox problems.

Concerning KMnO4, if you have never done it you should perform a chameleon reaction. Lots of youtube vids and basic to do. Dissolve 2.5 grams each of sugar and NaOH in 150mL of water. Add a small amount of relatively dilute KMnO4 solution (enough to make the 150mL a medium purple that you can still see through). Stir and watch.
You will see a slow reduction of the Mn from +7 through to +2 changing the solution from purple to blue to green to yellow and then, under ideal conditions the yellow fading and a slight pink appearing.
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[*] posted on 22-4-2015 at 07:02


Quote: Originally posted by byko3y  
Actually it's Mn(VI) and Mn(III), two most table states.
Chemistry is all about practice. You can't know nothing until you test it.
Reduction-oxidation potential can be measured, but they are different at different ph.


Permanganate is Mn(+7). The most 'stable' OS of Mn is Mn(+2). But 'stability' is a fairly meaningless term here: reducing permanganate will depend on conditions and the OS of the reduced Mn also depends on them (broadly speaking Mn(+2) in acid conditions, MnO2 (Mn(+4)) in neutral/alkaline conditions).

Not all standard reduction potentials are pH dependent.

For (proposed) redox reactions, determine the cell potential for the redox reaction from the oxidation and reduction half reactions: Ecell = Eox + Ered. Ecell must be positive (> 0) for the reaction to be thermodynamically feasible (but that still doesn't mean the reaction will actually proceed or is practical).

Selecting the right half reactions (oxidation and reduction) will tell you what are the reaction products.

[Edited on 22-4-2015 by blogfast25]




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byko3y
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[*] posted on 22-4-2015 at 08:04


Oh my godness, I'm so-so sorry, I actually was talking about chromium, but for some reason called it manganese. Was wondering "why can't I find Mn2(SO4)3, I know it should be somewhere here".
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blogfast25
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[*] posted on 22-4-2015 at 08:23


Quote: Originally posted by byko3y  
Oh my godness, I'm so-so sorry, I actually was talking about chromium, but for some reason called it manganese. Was wondering "why can't I find Mn2(SO4)3, I know it should be somewhere here".


Mn2(SO4)3 does exist but only in solution. Mn(+3) is in fact relatively stable but less compared to Mn(+7), Mn(+4) or Mn(+2).




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Madchemyst
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[*] posted on 22-4-2015 at 09:27


Thanks so much for all the reply's.. Blogfast25 that info was very helpful. I have been using the "All Lab no Lecture" book along with reading a text book. It has been many years since college. I am trying to understand what is going and not just doing experiments. I have been really enjoying my learning and this forum is really amazing....Thanks again all........
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blogfast25
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[*] posted on 22-4-2015 at 09:54


Quite a large table of SRPs (conveniently arranged in alphabetical order), here:

http://web.archive.org/web/20070518092613/http://www.northla...




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bolbol
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[*] posted on 22-4-2015 at 10:03


I was very confused about that a while ago as well, especially since I was messing with Nitric acid. I could solve the equations but I was always thinking to myself what makes the NO2 to NO ratio to be x, how can it be exact. And also about oxidation states. I would dissolve the non metals like Arsenic and wonder which oxidation state they would go to and if they happen to have 2-3 common oxidation states I'd wonder which ones they were oxidized to or I had doubts as to maybe only a certain amount were oxidized to +x and certain to +y. Then I read more about cell potentials and stuff like that and it cleared up just a bit for me but still I always lean towards the unknowns and spend hours confusing myself :D
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[*] posted on 22-4-2015 at 10:51


Quote: Originally posted by bolbol  
I was very confused about that a while ago as well, especially since I was messing with Nitric acid. I could solve the equations but I was always thinking to myself what makes the NO2 to NO ratio to be x, how can it be exact.
Is this a question?
Quote: Originally posted by bolbol  
but still I always lean towards the unknowns and spend hours confusing myself :D

Now you spend 2 minutes confusing me...




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bolbol
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[*] posted on 22-4-2015 at 12:54


oh no lol its not a question I was just saying what kind of stuff would confuse me.
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Molecular Manipulations
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[*] posted on 22-4-2015 at 13:01


So you know the answer now?



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[*] posted on 22-4-2015 at 14:09


I believe I do. I think since the H+ gets used up to form the water and a part of the new compound the left over NO3-'s N+5 is reduced forming either NO or NO2 depending on what the it is being reduced with and additionally the NO reacts with NO2 as it bubbles out of the reaction vessel.
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